Posted by Mike on Friday, April 9, 2010 at 5:24pm.
A certain 9-V battery measures 9.0 V with no load connected to it, but only 8.5 V when connected to a radio which draws 150 mA of current.
What is the internal resistance of the battery?
- college physics - bobpursley, Friday, April 9, 2010 at 5:46pm
- college physics - Mike, Friday, April 9, 2010 at 5:54pm
I see you're basically finding the voltage drop that occurs as a result of that internal resistance.
Now they ask, in the second part .... What would be the terminal voltage for a 50-W load?
- college physics - Mike, Friday, April 9, 2010 at 5:57pm
Oops. question should say 50 ohm load NOT 50-W load. Sorry.
Terminal Voltage as I understand it would be the Vt = Vo - I(Ri)
Where Vt is terminal voltage, Vo is battery voltage, I is current and Ri is internal resistance. Is that correct? How would they do it?
- college physics - bobpursley, Friday, April 9, 2010 at 6:01pm
current = 9.0/(internal resisitance + 50)
terminal voltage= 9.0-current*Ri
- college physics - Mike, Friday, April 9, 2010 at 6:07pm
thank you very much
- college physics - Mike, Friday, April 9, 2010 at 6:24pm
sorry here's another question ... they also ask what is the maximum amount of power you can get out of this battery?
You have P = IV. Max voltage is 9V, and max current is 2.7A. Then in that case that would be 9 * 2.7 = 24.3. Did I do that right?
- college physics - bobpursley, Friday, April 9, 2010 at 8:21pm
No, not at all. Max power is delived when load resistance is equal to internal resistance.
Proof: Load power= Rl*Il^2=Rl(9/(Ri+Rl)^2
take the derivative:
9/(Ri+Rl)^2 - 18Rl/(Ri+Rl)^3
set that equal to zero, and solve for Rl
Rl=Ri for max power.
- college physics - Damon, Friday, April 9, 2010 at 8:23pm
Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.
- college physics - Damon, Friday, April 9, 2010 at 8:26pm
However as far as useful power out of the battery instead of generating heat inside you had best look for where the external voltage times the current ins maximum.
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